# Are $121$ and $400$ the only perfect squares of the form $\sum\limits_{k=0}^{n}p^k$?

I've been looking for perfect squares that can be represented as $\sum\limits_{k=0}^{n}p^k$.

Of course, both $n$ and $p$ should be natural numbers larger than $1$.

Searching up to $n=100$ and $p=200$, I found only $2$ cases:

• $121=11^2=3^0+3^1+3^2+3^3+3^4=\sum\limits_{k=0}^{4}3^k$
• $400=20^2=7^0+7^1+7^2+7^3=\sum\limits_{k=0}^{3}7^k$

Is there any way to prove that there are no other cases?

• $p$ an integer or $p$ a prime? – Hagen von Eitzen Jan 26 '15 at 22:07
• Some of thorniest open problems in number theory come about from this interplay between additive number theory and multiplicative number theory. – James47 Jan 26 '15 at 22:35
• @James47: It reminds me of something I read on some "mathematical novel". I think it was The Music of Primes or Fermat's Last Theorem... Anyway, it said something like - "Mathematics is like a vast ocean of unknown, with little islands of knowledge and nothing that connects between them"... So if I understand you correctly, those two theories that you've mentioned are two separate islands in that ocean. – barak manos Jan 26 '15 at 22:43
• You can rewrite $\sum_{k=0}^n p^k = \frac{p^{k+1}-1}{p-1}$ – azimut Jan 26 '15 at 23:02
• Definitely of interest: arxiv.org/pdf/1312.4037v1.pdf – Barry Cipra Jan 27 '15 at 1:56

Turning a comment into an answer (of sorts)....

Various papers (e.g., this one by Yann Bugeaud and Preda Mihailescu) cite papers by Nagell and Ljunggren as proving there are no perfect squares of the given form other than the ones the OP found.

As for the more general problem of representing an arbitrary perfect power, a recent paper by Michael Bennett and Aaron Levin offers this assessment in its abstract (boldface emphasis added):

The Diophantine equation ${x^n−1\over x-1} = y^q$ has four known solutions in integers $x$, $y$, $q$ and $n$ with $|x|, |y|, q \gt 1$ and $n \gt 2$. Whilst we expect that there are, in fact, no more solutions, such a result is well beyond current technology.

The other two solutions referred to are the perfect cube $7^3$ expressed as $1+18+18^2$ and $1+(-19)+(-19)^2$.